AQA A-level Physics coverage

Medical physics (A-level only)

Section 3.10
19 spec leafs
Optional · choose 1 of 5

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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3.10.1.1

Physics of vision

  • The cornea supplies most of the eye's refraction; the lens changes power during accommodation so that a real image forms on the retina.
  • Cone cells work best in bright light, provide colour vision and give high spatial resolution because their nerve pathways show little convergence.
  • Rod cells are more sensitive at low light levels but do not distinguish colour; many rods share a nerve pathway, reducing spatial resolution.
  • The eye's spectral response depends on wavelength and on which receptors are active. A common error is to describe rods as producing detailed colour vision.

Tier 1 · Easy

2 marks
ORIGINAL

State the type of retinal receptor that is mainly responsible for vision at very low light intensity and give one property of the resulting image.

Tier 2 · Standard

3 marks
ORIGINAL

Explain why two faint points of light that fall on nearby rod cells may be seen as one point even though both points are detected.

Tier 3 · Hard

5 marks
ORIGINAL

A coloured grid is viewed first in bright light and then after the illumination is greatly reduced. Explain the changes expected in colour, detail and sensitivity, referring to the retinal receptors and their nerve connections.

3.10.1.2

Defects of vision and their correction using lenses

  • Lens power is P=1/fP=1/f with ff in metres and PP in dioptres: converging lenses have positive power and diverging lenses have negative power.
  • Using the real-is-positive convention, 1f=1u+1v\frac{1}{f}=\frac{1}{u}+\frac{1}{v}; a virtual object or image distance is negative, and m=v/um=v/u.
  • Myopia focuses distant objects in front of the retina and is corrected by a diverging lens; hypermetropia needs a converging lens to view near objects clearly.
  • Astigmatism is corrected with a cylindrical component whose prescription includes its power and an axis between 00^\circ and 180180^\circ.
  • Keep all distances in metres before finding power. A common error is to use the magnitude of a virtual image distance as positive.

Tier 1 · Easy

2 marks
ORIGINAL

State the type of spectacle lens used to correct myopia and state the sign of its power.

Tier 2 · Standard

3 marks
ORIGINAL

A converging lens has power +5.0D+5.0\,\text{D}. An object is 0.60m0.60\,\text{m} from the lens. Using the real-is-positive convention, calculate the image distance and the magnitude of the magnification.

Tier 3 · Hard

5 marks
ORIGINAL

A hypermetropic eye has a near point 0.80m0.80\,\text{m} from a spectacle lens. Determine the lens power needed so that an object 0.25m0.25\,\text{m} from the lens forms a virtual image at the near point. Use the real-is-positive convention.

3.10.2.1

Ear as a sound detection system

  • The pinna and ear canal direct sound to the tympanic membrane, whose vibrations move the ossicles in the middle ear.
  • The ossicles act as a lever system, and the smaller area of the oval window than the tympanic membrane produces a larger pressure in the cochlear fluid.
  • Pressure waves in the cochlea move regions of the basilar membrane; hair cells convert this mechanical motion into electrical signals in the auditory nerve.
  • When explaining transmission, follow the complete chain from pressure variation in air to mechanical vibration, fluid motion and an electrical nerve signal.

Tier 1 · Easy

2 marks
ORIGINAL

Name the membrane that vibrates when sound reaches the end of the ear canal and name the three-bone system moved by it.

Tier 2 · Standard

3 marks
ORIGINAL

Explain two features of the middle ear that increase the pressure delivered to the fluid in the cochlea.

Tier 3 · Hard

5 marks
ORIGINAL

A sound produces pressure amplitude 0.012Pa0.012\,\text{Pa} on a tympanic membrane of area 5.5×105m25.5\times10^{-5}\,\text{m}^2. The ossicles increase the force by a factor of 1.41.4 and act on an oval window of area 3.2×106m23.2\times10^{-6}\,\text{m}^2. Calculate the pressure amplitude delivered to the inner-ear fluid.

3.10.2.2

Sensitivity and frequency response

  • Sound intensity is power transferred per unit area, measured in W m2\text{W m}^{-2}; for an isotropic point source, I=P/(4πr2)I=P/(4\pi r^2).
  • Intensity level is L=10log10(I/I0)L=10\log_{10}(I/I_0) in decibels, where I0=1.0×1012W m2I_0=1.0\times10^{-12}\,\text{W m}^{-2}.
  • Equal-loudness curves show the intensity level needed at each frequency for the same perceived loudness; normal hearing is most sensitive around a few kilohertz.
  • The logarithmic decibel scale reflects human perception: a change in intensity level is found from ΔL=10log10(I2/I1)\Delta L=10\log_{10}(I_2/I_1), not from an intensity ratio in decibels.
  • The dBA scale applies frequency weighting to approximate the response of human hearing; do not confuse dBA weighting with an unweighted dB intensity level.

Tier 1 · Easy

2 marks
ORIGINAL

Calculate the intensity level of a sound with intensity 1.0×108W m21.0\times10^{-8}\,\text{W m}^{-2}. Use I0=1.0×1012W m2I_0=1.0\times10^{-12}\,\text{W m}^{-2}.

Tier 2 · Standard

3 marks
ORIGINAL

Two sounds have intensity levels 76dB76\,\text{dB} and 58dB58\,\text{dB}. Determine the ratio of the larger intensity to the smaller intensity.

Tier 3 · Hard

5 marks
ORIGINAL

An isotropic sound source has power 0.80W0.80\,\text{W}. Determine the distance at which its intensity level is 85dB85\,\text{dB}. Then explain why a 12kHz12\,\text{kHz} tone at this level may sound quieter than a 3kHz3\,\text{kHz} tone at the same level. Use I0=1.0×1012W m2I_0=1.0\times10^{-12}\,\text{W m}^{-2}.

3.10.2.3

Defects of hearing

  • Hearing loss raises the intensity level required for a sound to be heard, so the affected equal-loudness or threshold curve shifts upward over the damaged frequency range.
  • Age-related deterioration is typically greatest at high frequencies, whereas prolonged excessive noise exposure commonly produces the greatest loss near 4kHz4\,\text{kHz}.
  • A hearing loss in decibels is the vertical difference between normal and impaired threshold curves at the stated frequency.
  • The specification tests changes in hearing and equal-loudness curves, not the physiological mechanism of the damage; avoid unsupported descriptions of structures being damaged.

Tier 1 · Easy

1 mark
ORIGINAL

State the frequency range in which age-related hearing deterioration is usually greatest.

Tier 2 · Standard

3 marks
ORIGINAL

At 4.0kHz4.0\,\text{kHz}, a normal threshold is 8dB8\,\text{dB} and an impaired threshold is 38dB38\,\text{dB}. Calculate the hearing loss and the factor by which the threshold intensity has increased.

Tier 3 · Hard

5 marks
ORIGINAL

Listener A has hearing losses of 6dB6\,\text{dB} at 0.5kHz0.5\,\text{kHz}, 15dB15\,\text{dB} at 4kHz4\,\text{kHz} and 44dB44\,\text{dB} at 10kHz10\,\text{kHz}. Listener B has losses of 9dB9\,\text{dB}, 48dB48\,\text{dB} and 17dB17\,\text{dB} at the same frequencies. Deduce the likely cause for each pattern and explain how each pattern changes an equal-loudness curve and perceived sound.

3.10.3.1

Simple ECG machines and the normal ECG waveform

  • Skin electrodes measure small potential differences produced by the heart; conductive gel, clean skin and secure electrodes reduce contact resistance and movement artefacts.
  • A differential, high-gain, low-noise amplifier and shielded leads are used because the ECG signal is only of order millivolts and is vulnerable to electrical interference.
  • The P wave represents atrial depolarisation, the QRS complex ventricular depolarisation, and the T wave ventricular repolarisation.
  • Heart rate is found from successive equivalent points, normally R peaks: rate=60/T\text{rate}=60/T when the period TT is in seconds. Measure several cycles to reduce percentage uncertainty.
  • Do not describe the ECG as a direct trace of blood pressure or mechanical force; it records electrical potential difference.

Tier 1 · Easy

2 marks
ORIGINAL

Identify the electrical events represented by the QRS complex and the T wave in a normal ECG.

Tier 2 · Standard

3 marks
ORIGINAL

Five successive R peaks occur at 0.42s0.42\,\text{s}, 1.20s1.20\,\text{s}, 1.98s1.98\,\text{s}, 2.76s2.76\,\text{s} and 3.54s3.54\,\text{s}. Determine the mean heart rate in beats per minute.

Tier 3 · Hard

5 marks
ORIGINAL

An ECG trace contains slow baseline changes when the patient moves and a regular interference signal from nearby mains equipment. Explain how electrode attachment, the leads and the amplifier should be arranged to obtain a clearer normal waveform.

3.10.4.1

Ultrasound imaging

  • Acoustic impedance is Z=ρcZ=\rho c. At a boundary, the reflected intensity fraction is Ir/Ii=((Z2Z1)/(Z2+Z1))2I_r/I_i=((Z_2-Z_1)/(Z_2+Z_1))^2.
  • Coupling gel removes the air layer between transducer and skin, reducing the impedance mismatch and allowing more ultrasound to enter the body.
  • A piezoelectric crystal driven by a short alternating potential difference produces an ultrasound pulse; a returning echo deforms the crystal and generates a potential difference.
  • Echo depth follows d=ct/2d=ct/2 because the measured time includes the outward and return journeys. Greater frequency improves resolution but generally increases attenuation.
  • An A-scan plots echo amplitude against time or depth, whereas a B-scan uses echo brightness and position to build a two-dimensional image.
  • Ultrasound is non-ionising and can image moving structures in real time, but it has lower resolution than some alternatives and is strongly reflected at air or bone boundaries.

Tier 1 · Easy

2 marks
ORIGINAL

A display shows echo amplitude against time rather than a two-dimensional image. Identify the ultrasound scan type and state what a B-scan displays instead.

Tier 2 · Standard

4 marks
ORIGINAL

Ultrasound passes normally from tissue A, where ρ=1000kg m3\rho=1000\,\text{kg m}^{-3} and c=1500m s1c=1500\,\text{m s}^{-1}, into tissue B, where ρ=1060kg m3\rho=1060\,\text{kg m}^{-3} and c=1540m s1c=1540\,\text{m s}^{-1}. Calculate the percentage of incident intensity reflected at the boundary.

Tier 3 · Hard

6 marks
ORIGINAL

A 4.0MHz4.0\,\text{MHz} ultrasound pulse travels through tissue at 1540m s11540\,\text{m s}^{-1} and returns to the transducer 130μs130\,\mu\text{s} after transmission. Calculate the reflector depth and estimate the scan resolution as one wavelength. Explain why coupling gel is placed between the transducer and skin.

3.10.4.2

Fibre optics and endoscopy

  • An optical fibre has a higher-index core surrounded by lower-index cladding, so rays incident above the critical angle undergo total internal reflection at the interface.
  • A coherent bundle keeps the fibres in the same relative positions at both ends and transfers an image; cladding prevents light crossing between fibres and blurring that image.
  • A non-coherent bundle does not preserve fibre order and carries illumination into the body, so its individual fibre positions do not encode an image.
  • Flexible fibre bundles permit internal imaging through a small opening and can guide illumination or treatment light with less invasive access.
  • For total internal reflection, light must travel from higher to lower refractive index and the incidence angle must exceed the critical angle; merely stating that light reflects is insufficient.

Tier 1 · Easy

2 marks
ORIGINAL

Name the endoscope fibre bundle that carries an image and the bundle that carries illumination into the body.

Tier 2 · Standard

3 marks
ORIGINAL

A fibre has core refractive index 1.621.62 and cladding refractive index 1.501.50. Calculate the critical angle at the core-cladding boundary and determine whether a ray incident there at 72.072.0^\circ undergoes total internal reflection.

Tier 3 · Hard

5 marks
ORIGINAL

A damaged endoscope has an image bundle whose fibre positions are rearranged between its two ends, and some cladding has been removed. Explain the effects on the observed image and why the illumination bundle can still work when its fibre order is random.

3.10.4.3

Magnetic resonance (MR) scanner

  • A strong static magnetic field, produced by a superconducting magnet, makes hydrogen nuclei (protons) adopt aligned spin states and precess about the field direction; the field does not make stationary protons start spinning classically.
  • A short radio-frequency pulse at the appropriate frequency excites protons and changes their spin state in the selected region.
  • After the pulse, excited protons relax or de-excite and emit radio-frequency signals that receiver coils detect; detailed relaxation-time calculations are not required.
  • Gradient magnetic fields vary with position so successive small regions of a cross-section can be selected and located.
  • A computer processes the detected signals to construct a cross-sectional image. MR uses no ionising radiation and is particularly useful for soft-tissue contrast.

Tier 1 · Easy

2 marks
ORIGINAL

State which nuclei provide the main signal in an MR scanner and describe their behaviour in the scanner's static magnetic field.

Tier 2 · Standard

3 marks
ORIGINAL

Describe what happens to selected protons during and immediately after a short radio-frequency pulse in an MR scan.

Tier 3 · Hard

5 marks
ORIGINAL

Outline how an MR scanner obtains a cross-sectional image, beginning with a patient in the main magnetic field and ending with a computer-generated image.

3.10.5.1

The physics of diagnostic X-rays

  • Electrons released by thermionic emission are accelerated through a potential difference VV and decelerated at a metal target. Most input energy heats the target; a rotating anode spreads this heating over a larger area.
  • The continuous spectrum comes from electrons losing different amounts of kinetic energy. The maximum photon energy is Emax=eVE_{\max}=eV, so the minimum wavelength is λmin=hc/(eV)\lambda_{\min}=hc/(eV).
  • Characteristic lines arise when incident electrons remove inner-shell electrons and higher-shell electrons fall into the vacancies; their photon energies are differences between atomic energy levels.
  • Increasing tube current increases beam intensity and patient dose, whereas increasing accelerating voltage raises the maximum photon energy and penetration. A smaller focal spot improves sharpness but concentrates heating; filtration and collimation remove low-energy photons and restrict the exposed region.

Tier 1 · Easy

1 mark
ORIGINAL

State the maximum photon energy, in keV\text{keV}, from an X-ray tube operating at 68kV68\,\text{kV}.

Tier 2 · Standard

3 marks
ORIGINAL

An X-ray tube uses an accelerating potential difference of 72kV72\,\text{kV}. Calculate the minimum X-ray wavelength. Use e=1.60×1019Ce=1.60\times10^{-19}\,\text{C}, h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s} and c=3.00×108m s1c=3.00\times10^8\,\text{m s}^{-1}.

Tier 3 · Hard

5 marks
ORIGINAL

A rotating-anode tube operates at 95kV95\,\text{kV} with a current of 3.0mA3.0\,\text{mA} for 0.080s0.080\,\text{s}. It converts 1.2%1.2\% of the electrical energy into X-rays. Calculate the maximum photon energy and the total X-ray energy produced. Explain one benefit of rotating the anode. Use e=1.60×1019Ce=1.60\times10^{-19}\,\text{C}.

3.10.5.2

Image detection and enhancement

  • In a flat-panel detector, a scintillator converts X-rays to visible photons, photodiode pixels convert the light to electrical charge, and electronic scanning reads the pixel signals into a computer.
  • Flat-panel detectors are sensitive, produce an immediate digital image and allow computer enhancement, storage and transmission. Their wide dynamic range can reduce repeat exposures compared with photographic film.
  • X-rays darken photographic film. An intensifying screen converts X-rays to visible light so that less X-ray exposure is needed; fluoroscopic image intensification gives a live image but prolonged viewing can increase patient dose.
  • An X-ray-opaque contrast medium such as barium absorbs strongly. It outlines a low-contrast organ against surrounding soft tissue; a common error is to describe the image as being formed by reflected X-rays.

Tier 1 · Easy

1 mark
ORIGINAL

State the function of the scintillator in a flat-panel X-ray detector.

Tier 2 · Standard

3 marks
ORIGINAL

Explain why a patient is given a barium suspension before an X-ray image of the digestive tract is recorded.

Tier 3 · Hard

5 marks
ORIGINAL

A clinic needs a single abdominal X-ray image that can be enhanced immediately while keeping patient dose as low as practicable. Discuss the suitability of photographic film, fluoroscopic image intensification and a flat-panel detector, and justify a choice.

3.10.5.3

Absorption of X-rays

  • For a narrow monoenergetic beam, transmitted intensity follows I=I0eμxI=I_0e^{-\mu x}, where μ\mu is the linear attenuation coefficient and xx is the material thickness. Use compatible units for μ\mu and xx.
  • The half-value thickness is x1/2=ln2/μx_{1/2}=\ln 2/\mu. After nn half-value thicknesses the intensity is I0(1/2)nI_0(1/2)^n; do not treat attenuation as a constant subtraction per unit thickness.
  • The mass attenuation coefficient is μm=μ/ρ\mu_m=\mu/\rho and has units m2kg1\text{m}^2\,\text{kg}^{-1} when SI units are used. It allows attenuation properties to be compared without the direct effect of density.
  • Bone and contrast media generally attenuate X-rays more strongly than soft tissue, producing differential detector signals. Examiners require transmitted, not absorbed, intensity when I=I0eμxI=I_0e^{-\mu x} is used.

Tier 1 · Easy

2 marks
ORIGINAL

An absorber has linear attenuation coefficient 0.28cm10.28\,\text{cm}^{-1}. Calculate its half-value thickness.

Tier 2 · Standard

3 marks
ORIGINAL

A narrow X-ray beam of intensity 6.4W m26.4\,\text{W m}^{-2} passes through 7.0mm7.0\,\text{mm} of tissue with μ=0.18mm1\mu=0.18\,\text{mm}^{-1}. Calculate the transmitted intensity.

Tier 3 · Hard

5 marks
ORIGINAL

An incident X-ray beam has intensity 10.0W m210.0\,\text{W m}^{-2}. One ray crosses 18mm18\,\text{mm} of soft tissue with μ=0.050mm1\mu=0.050\,\text{mm}^{-1} and then 4.0mm4.0\,\text{mm} of bone with μ=0.32mm1\mu=0.32\,\text{mm}^{-1}. A neighbouring ray crosses 22mm22\,\text{mm} of the same soft tissue only. Calculate both transmitted intensities and the ratio of the soft-tissue-only intensity to the bone-path intensity.

3.10.5.4

CT scanner

  • A CT scanner moves an X-ray tube around the patient and sends a narrow, approximately monochromatic beam through the body to an array of detectors at many angles.
  • Each detector measures transmitted intensity along a path. A computer combines many projections to reconstruct cross-sectional slices and can assemble successive slices into a three-dimensional representation.
  • CT removes the superposition of structures found in a plain radiograph and can give improved contrast and spatial information, especially for internal soft-tissue structures.
  • CT is more expensive and usually gives a larger ionising-radiation dose than a simple X-ray image. Comparisons must link resolution, cost and safety to the clinical task rather than merely list scanner features.

Tier 1 · Easy

1 mark
ORIGINAL

State why the X-ray tube moves around a patient during a CT scan.

Tier 2 · Standard

3 marks
ORIGINAL

Describe how a CT scanner produces a cross-sectional image from X-rays transmitted through a patient.

Tier 3 · Hard

6 marks
ORIGINAL

A hospital is choosing between CT and a plain X-ray image to localise a small lung lesion partly hidden by overlapping ribs, and to confirm the position of a radiopaque feeding tube. Compare the techniques using image resolution, superposition, radiation dose, cost and availability. Recommend one technique for each task.

3.10.6.1

Imaging techniques

  • A tracer contains a gamma-emitting radioisotope attached to a compound with an affinity for a particular organ. Gamma radiation is penetrating enough to leave the body and be detected externally.
  • Technetium-99m has a physical half-life of about 6h6\,\text{h} and emits a useful 140keV140\,\text{keV} gamma photon. Its short half-life limits dose while allowing a scan, and it can be labelled to many biologically active compounds.
  • Iodine-131 has a half-life of about 8d8\,\text{d}, emits beta radiation and a principal 364keV364\,\text{keV} gamma photon, and is naturally taken up by the thyroid. Indium-111 has a half-life of about 2.8d2.8\,\text{d}, emits 171keV171\,\text{keV} and 245keV245\,\text{keV} gamma photons, and can label compounds or blood cells for longer investigations.
  • A molybdenum-technetium generator contains longer-lived molybdenum-99, whose decay produces technetium-99m. It lets hospitals obtain fresh tracer without a nearby reactor or accelerator.
  • In PET, a positron-emitting tracer produces annihilation photons. Two 511keV511\,\text{keV} gamma photons travelling in nearly opposite directions are detected in coincidence, allowing the computer to locate tracer activity.

Tier 1 · Easy

1 mark
ORIGINAL

State why a medical tracer used for external imaging should emit gamma radiation.

Tier 2 · Standard

3 marks
ORIGINAL

Justify the use of technetium-99m for imaging an organ. Refer to its 6h6\,\text{h} half-life, its 140keV140\,\text{keV} gamma emission and its chemical use.

Tier 3 · Hard

5 marks
ORIGINAL

Explain how a PET scan can map regions of high glucose uptake after a patient receives a positron-emitting glucose analogue. Include the nuclear event, the detection method and how position is inferred.

3.10.6.2

Half-life

  • Physical half-life TPT_P is the time for half the unstable nuclei to decay. It is a property of the radionuclide and is not altered by biological processes.
  • Biological half-life TBT_B is the time for biological removal processes to reduce the amount of tracer in an organ or body to half its value, ignoring radioactive decay.
  • Effective half-life TET_E describes the combined fall due to both processes and obeys 1/TE=1/TB+1/TP1/T_E=1/T_B+1/T_P. Consequently TET_E must be shorter than both TBT_B and TPT_P.
  • Keep all three half-lives in the same units before using the reciprocal equation. A common error is to add the half-lives themselves rather than their reciprocals.

Tier 1 · Easy

2 marks
ORIGINAL

Define the biological half-life of a tracer in an organ.

Tier 2 · Standard

3 marks
ORIGINAL

A tracer has physical half-life 18h18\,\text{h} and biological half-life 30h30\,\text{h}. Calculate its effective half-life.

Tier 3 · Hard

5 marks
ORIGINAL

A radionuclide has physical half-life 12.0h12.0\,\text{h}. Measurements in a patient give an effective half-life of 7.20h7.20\,\text{h}. Calculate the biological half-life and the percentage of the initial activity remaining in the patient after 21.6h21.6\,\text{h}.

3.10.6.3

Gamma camera

  • A lead collimator absorbs photons travelling in unsuitable directions, so accepted gamma photons carry directional information. Narrower holes improve spatial resolution but reduce count rate and sensitivity.
  • A scintillation crystal converts an absorbed gamma photon into a flash of visible light. A light guide spreads the flash to an array of photomultiplier tubes.
  • In a photomultiplier tube, light ejects photoelectrons from a photocathode. Successive dynodes at increasing potentials produce secondary emission, giving a large electron pulse at the anode.
  • The relative photomultiplier outputs locate each scintillation, while pulse size estimates photon energy so scattered photons can be rejected. A computer stores many accepted positions to build the image.

Tier 1 · Easy

1 mark
ORIGINAL

State the function of the lead collimator in a gamma camera.

Tier 2 · Standard

3 marks
ORIGINAL

Describe how a gamma photon entering a gamma camera produces an electrical pulse at the output of a photomultiplier tube.

Tier 3 · Hard

5 marks
ORIGINAL

Explain how a gamma camera determines the position of tracer activity and improves the quality of the recorded image. Include the roles of the collimator, photomultiplier array and pulse-height selection.

3.10.6.4

Use of high-energy X-rays

  • External radiotherapy directs high-energy X-rays into a tumour. Their penetration lets energy be delivered at depth, where ionisation damages DNA and prevents cells from dividing successfully.
  • Several beam directions can converge on the tumour. The tumour receives the full combined dose, while each region of healthy tissue lies in fewer beam paths and receives a smaller dose.
  • Collimation, shielding and computer planning restrict the irradiated volume. Imaging is used to locate the tumour before selecting beam directions and field shapes.
  • Treatment is divided into fractions so healthy cells can repair between exposures while the prescribed tumour dose accumulates. Do not claim that healthy tissue receives zero dose.

Tier 1 · Easy

1 mark
ORIGINAL

State why high-energy X-rays are used to treat a tumour deep inside the body.

Tier 2 · Standard

3 marks
ORIGINAL

Explain how directing X-ray beams at a tumour from several angles can reduce damage to healthy tissue.

Tier 3 · Hard

5 marks
ORIGINAL

A tumour of mass 0.18kg0.18\,\text{kg} receives 1.8Gy1.8\,\text{Gy} in each of 2020 X-ray treatment fractions. Calculate the total energy absorbed by the tumour. Explain two methods, other than reducing the prescribed tumour dose, that limit exposure of healthy cells.

3.10.6.5

Use of radioactive implants

  • Brachytherapy places sealed radioactive sources inside or very close to a tumour, giving a high local dose while reducing irradiation of more distant tissue.
  • Beta-emitting implants are useful because beta particles ionise tissue over a short range. Gamma radiation would penetrate farther and expose more healthy tissue outside the target region.
  • Several small sources can distribute dose through an irregular tumour. Source activity, half-life, position and treatment time determine the delivered dose; temporary implants can be removed after the planned exposure.
  • Staff exposure is reduced using handling tools, shielding, short handling times and increased distance where appropriate. A common weak answer states only that beta is less penetrating without linking this to localisation of dose.

Tier 1 · Easy

1 mark
ORIGINAL

State why a beta emitter is suitable for a radioactive implant placed inside a tumour.

Tier 2 · Standard

3 marks
ORIGINAL

Explain why a beta-emitting implant may expose less healthy tissue than an external high-energy X-ray beam used to deliver the same tumour dose.

Tier 3 · Hard

5 marks
ORIGINAL

A temporary beta-emitting implant has initial activity 9.6MBq9.6\,\text{MBq} and physical half-life 48h48\,\text{h}. Calculate its activity after 120h120\,\text{h}. Explain how the position and radiation type of the implant help protect healthy tissue.

3.10.6.6

Imaging comparisons

  • A plain X-ray image is fast, widely available and gives good spatial detail for bone, but it uses ionising radiation, gives a two-dimensional projection and has poor soft-tissue contrast.
  • CT gives cross-sectional images with better localisation and soft-tissue information than a plain radiograph, but it usually has a higher ionising dose, higher cost and lower availability.
  • Ultrasound is portable, relatively inexpensive, real-time and non-ionising, but image quality is operator-dependent and transmission through bone or gas is poor. MR gives excellent soft-tissue contrast without ionising radiation but is slower, expensive and less available.
  • Radionuclide and PET images show physiological function rather than only anatomy, but require an internal ionising tracer and usually have lower spatial resolution. A comparison must apply resolution, dose or safety, convenience or availability, and tissue contrast to the stated clinical need.

Tier 1 · Easy

1 mark
ORIGINAL

Select a suitable imaging technique for repeated bedside monitoring of moving soft tissue when ionising radiation must be avoided.

Tier 2 · Standard

3 marks
ORIGINAL

Compare CT and MR for repeated imaging of a brain tumour. Refer to soft-tissue contrast, patient dose and convenience.

Tier 3 · Hard

6 marks
ORIGINAL

A patient may have a kidney stone. Evaluate ultrasound, CT and MR for locating the stone, using spatial resolution, soft-tissue or stone contrast, radiation dose, availability and convenience. Reach a justified recommendation.